Question

1. A crackpot inventor designs the following heat engine. (An annotated PV plot might help.) Step D: Starting at State A (PA, VA, and TA25.0 °C), the gas undergoes a reversible and isothermal expansion until it reaches State B (PB = 1.000 atm, VB, and TB). Step : A dry ice / acetone bath is used to isochorically bring the gas to State C (Pc, Vc, and Tc -78.5 °C) Step : The gas undergoes a reversible adiabatic compression until it is brought back to its original state (PA, VA, and TA 25.0 °C). The working gas is exactly 1 mole of an ideally-behaving gas with a temperature- independent y = Cp/Cv$/3. The steps are repeated, and each cycle converts heat energy into work energy. Determine Cy and Cp of the gas (each in terms of R). a. b. Calculate the state variables PA, VA, VB, TB, Pc, and Vc. (Hints: Whenever possible, just use the definitions of isotherm and isochore. Then whenever possible use the ideal gas law. At some point, you'll need to use the adiabatic step to calculate PA or VA) c. Calculate (or in some cases just determine) w, q, AU, and AH (not necessarily in that order) to the nearest J for . Step D; ii. Step ; and iii. Step

Answer 1

② Now using Isothermal change Propertig we have Piva = PRVB PAVA = 1.000X.24.46 PAVA = 24.40 - I S Using Adiabatic change rule we chave. PAVAS = Pero e PAVAT = 0.65 X (24.467 eph = and 8 = 5 PAVA 24.46 PA VAS - 0.65 x 84.4678 VAY = 60 the X (24.467-4 VA8r1 = 0.65 (24.468 just taking inverse calulation 8= 55. ie. VA = 18.354 L doing - and using Nowe PAVA = 24.46 PA = 24.46 = 1.33.atm 18.354. Thus finally we have PA = 1.33 atm ; VA = 18.354 L Vc = 24.46L ; Pc = 0.65 atm V8 = 24.46 L ; TB = 298 K Scanned with CamScanner

c> for is step o lie. Isothermoel reversible ) as W= n Rien va . ; R= 8.314 J mostky W = 1X8.314x298 x en 24:46 18.854 WE 2477.57 x 0.2871 W = 711.38 J . cause DT=0) by. AU=0 C for isothermal process & au= ncrot = ax C x 0 . c7 Au=q+w Here q=-W - - 711.38 J py for slep 2 ise. Isochor's [Ov=o] ax W = Pov=0 > Чат Ср БТ = 1X3 X 8.314 x C194.5'- 2983 2 = -1,290.7 J c) AU = qtw = qto = -1,290:7) cs Scanned with CamScanner

3) slep 3. for cediabatic process as Here q=0 (NO cheat exchanges) by W=-[ PAVA - Pavad F- S 11.33x18.354 - 0.65X24:46] 8.314 L =-1.535 J c) su=htw AU -1535 J Note: We have taken R do 8.314 cause we want our the modynamic properties in Joules. Scanned with CamScanner

9.1 Ans : ay Here 1st we have to determine cp and cu - auch that we have given 8=co/cr. . & no. of moles = 1 *The ep for duch a cystem is given by the formula cp= 5 nr. e. Jcp = 55 R cause n=1 in our case. * Similarly er fors such a system is given by - the formula cv=BnR lie. Cv- rl cause nat. by Now we have to find out the state variables PA , VA, VB To, pc and Vc. Given : TA = 25.0°C = 298 K TC = -78.5°C = 194'5K & PB = 1.000 atm * we have provided the information regarding the state variables going from & state to other for that we can plot a graphical nature as 9 CPA, VA,TA) → Iso thermal reversible expansion @+isochovic @=> reversible adibatic compression. Pc, Vc. To 07 CP VO, Ie) Scanned With CamScanter

* Using fact for Isothermal Expansion → For it bt- constant , i.e. TA = Te → TA = Tg = 298 K How To find VB . we have PB = 1.000 atm, To= 298 K Using ideal gas equation PB VB = nRTB ; R=0.0821 Latm molky n1 mol. VB - n RTB = 1X0.082 x 2.98 = 24.46 L = Ve I * Now from State (PB, VB, TB) state CPC, Vc, Tc) we have i sochoric behaviour lie. sv = constant, i.e. Ve=Ve → VBFV = 24:46 L • To calculate Pe use ideal gas law Perc=MRTC ;. Te = 194.5k R=0.0821 Latinamotky 1. NRTC Pc = Vc knal - 1X0:0821X194.5 24.46 Pc = 0.65 atm out PA and VA .. to find as Scanned with Now lets try o CamScanner